Calculus/Limits/Exercises

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	Limits/Exercises

Basic Limit ExercisesEdit

1.

\lim _{x\to 2}{\Big [}4x^{2}-3x-1{\Big ]}

9

9

2.

\lim _{x\to 5}{\Big [}x^{2}{\Big ]}

25

25

3.

\lim _{x\to {\frac {\pi }{4}}}{\Big [}\cos ^{2}(x){\Big ]}

1/2

1/2

4.

\lim _{x\to 1}{\Big [}5e^{x-1}-5{\Big ]}

0

0

Solutions

One-Sided LimitsEdit

Evaluate the following limits or state that the limit does not exist.

5.

\lim _{x\to 0^{-}}{\frac {x^{3}+x^{2}}{x^{3}+2x^{2}}}

{\frac {1}{2}}

{\frac {1}{2}}

6.

\lim _{x\to 7^{-}}{\Big [}|x^{2}+x|-x{\Big ]}

49

49

7.

\lim _{x\to -1^{+}}{\sqrt {1-x^{2}}}

0

0

8.

\lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}

The limit does not exist.

9.

\lim _{x\to 0^{-}}{\frac {|x|}{x}}

-1

-1

10.

\lim _{x\to 1^{-}}\arcsin(x)

{\frac {\pi }{2}}

{\frac {\pi }{2}}

Solutions

Two-Sided LimitsEdit

Evaluate the following limits or state that the limit does not exist.

11.

\lim _{x\to -1}{\frac {1}{x-1}}

-{\frac {1}{2}}

-{\frac {1}{2}}

12.

\lim _{x\to 4}{\frac {1}{x-4}}

The limit does not exist.

13.

\lim _{x\to 2}{\frac {1}{x-2}}

The limit does not exist.

14.

\lim _{x\to -3}{\frac {x^{2}-9}{x+3}}

-6

-6

15.

\lim _{x\to 3}{\frac {x^{2}-9}{x-3}}

6

6

16.

\lim _{x\to -1}{\frac {x^{2}+2x+1}{x+1}}

0

0

17.

\lim _{x\to -1}{\frac {x^{3}+1}{x+1}}

3

3

18.

\lim _{x\to 4}{\frac {x^{2}+5x-36}{x^{2}-16}}

{\frac {13}{8}}

{\frac {13}{8}}

19.

\lim _{x\to 25}{\frac {x-25}{{\sqrt {x}}-5}}

10

10

20.

\lim _{x\to 0}{\frac {|x|}{x}}

The limit does not exist.

21.

\lim _{x\to 2}{\frac {1}{(x-2)^{2}}}

\infty

\infty

22.

\lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}

The limit does not exist.

23.

\lim _{x\to -2}{\frac {3x^{2}-8x-3}{2x^{2}-18}}

-{\frac {5}{2}}

-{\frac {5}{2}}

24.

\lim _{x\to 2}{\frac {x^{2}+2x+1}{x^{2}-2x+1}}

9

9

25.

\lim _{x\to 3}{\frac {x+3}{x^{2}-9}}

The limit does not exist.

26.

\lim _{x\to -1}{\frac {x+1}{x^{2}+x}}

-1

-1

27.

\lim _{x\to 1}{\frac {1}{x^{2}+1}}

{\frac {1}{2}}

{\frac {1}{2}}

28.

\lim _{x\to 1}\left[x^{2}+5x-{\frac {1}{2-x}}\right]

5

5

29.

\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}

0

0

30.

\lim _{x\to 1}{\frac {x^{2}-1}{x^{2}+2x-3}}

{\frac {1}{2}}

{\frac {1}{2}}

31.

\lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}

The limit does not exist.

32.

\lim _{x\to -1}\ln({\sqrt {x^{2}-1}})

The limit does not exist.

33.

\lim _{x\to 1}\arcsin(x)

The limit does not exist.

Solutions

Limits to InfinityEdit

Evaluate the following limits or state that the limit does not exist.

34.

\lim _{x\to \infty }{\frac {-x+\pi }{x^{2}+3x+2}}

0

0

35.

\lim _{x\to -\infty }{\frac {x^{2}+2x+1}{3x^{2}+1}}

{\frac {1}{3}}

{\frac {1}{3}}

36.

\lim _{x\to -\infty }{\frac {3x^{2}+x}{2x^{2}-15}}

{\frac {3}{2}}

{\frac {3}{2}}

37.

\lim _{x\to -\infty }{\Big [}3x^{2}-2x+1{\Big ]}

\infty

\infty

38.

\lim _{x\to \infty }{\frac {2x^{2}-32}{x^{3}-64}}

0

0

39.

\lim _{x\to \infty }6

6

6

40.

\lim _{x\to \infty }{\frac {3x^{2}+4x}{x^{4}+2}}

0

0

41.

\lim _{x\to -\infty }{\frac {2x+3x^{2}+1}{2x^{2}+3}}

{\frac {3}{2}}

{\frac {3}{2}}

42.

\lim _{x\to -\infty }{\frac {x^{3}-3x^{2}+1}{3x^{2}+x+5}}

-\infty

-\infty

43.

\lim _{x\to \infty }{\frac {x^{2}+2}{x^{3}-2}}

0

0

44.

\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}

0

0

45.

\lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)

\infty

\infty

46.

\lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}

-{\frac {2}{\pi }}

-{\frac {2}{\pi }}

Solutions

Limits of Piecewise FunctionsEdit

Evaluate the following limits or state that the limit does not exist.

48. Consider the function

f(x)={\begin{cases}(x-2)^{2}&{\text{if }}x<2\\x-3&{\text{if }}x\geq 2\end{cases}}

a.

\lim _{x\to 2^{-}}f(x)

0

0

b.

\lim _{x\to 2^{+}}f(x)

-1

-1

c.

\lim _{x\to 2}f(x)

The limit does not exist

49. Consider the function

g(x)={\begin{cases}-2x+1&{\text{if }}x\leq 0\\x+1&{\text{if }}0<x<4\\x^{2}+2&{\text{if }}x\geq 4\end{cases}}

a.

\lim _{x\to 4^{+}}g(x)

18

18

b.

\lim _{x\to 4^{-}}g(x)

5

5

c.

\lim _{x\to 0^{+}}g(x)

1

1

d.

\lim _{x\to 0^{-}}g(x)

1

1

e.

\lim _{x\to 0}g(x)

1

1

f.

\lim _{x\to 1}g(x)

2

2

50. Consider the function

h(x)={\begin{cases}2x-3&{\text{if }}x<2\\8&{\text{if }}x=2\\-x+3&{\text{if }}x>2\end{cases}}

a.

\lim _{x\to 0}h(x)

-3

-3

b.

\lim _{x\to 2^{-}}h(x)

1

1

c.

\lim _{x\to 2^{+}}h(x)

1

1

d.

\lim _{x\to 2}h(x)

1

1

Solutions

Intermediate Value TheoremEdit

51. Use the intermediate value theorem to show that there exists a value

f(x)=3

for

f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}

from

1\leq x\leq 9

. If you cannot use the intermediate value theorem to show this, explain why.

Notice

f(x)

is continuous from

x\in \left[1,9\right]

. Ergo, the intermediate value theorem applies. For all

y\in \left[2,{\frac {10}{3}}\right]

, there exists a

c\in (1,9)

so that

f(c)=3

.

\blacksquare

Notice

f(x)

is continuous from

x\in \left[1,9\right]

. Ergo, the intermediate value theorem applies. For all

y\in \left[2,{\frac {10}{3}}\right]

, there exists a

c\in (1,9)

so that

f(c)=3

.

\blacksquare

52. Use the intermediate value theorem to show that there exists an

x=\mu

so that

f(\mu )={\frac {\pi }{4}}

for

f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}

from

\pi \leq x\leq {\frac {9}{4}}\pi

. If you cannot use the intermediate value theorem to show this, explain why.

Notice

f(x)

is continuous from

x\in \left[\pi ,{\frac {9}{4}}\pi \right]

. Ergo, the intermediate value theorem applies.

f(\pi )={\frac {1}{\ln(\pi )}}

f\left({\frac {9}{4}}\pi \right)=0

It is known the following is true: $1\geq {\frac {\pi }{4}}\geq 0$ . From there, we can directly argue the following:

{\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}

By the intermediate value theorem, if

f(x)

is continuous from

x\in \left[\pi ,{\frac {9}{4}}\pi \right]

, then there exists an

x=\mu

so that

f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )

for

\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]

.

\blacksquare

Notice

f(x)

is continuous from

x\in \left[\pi ,{\frac {9}{4}}\pi \right]

. Ergo, the intermediate value theorem applies.

f(\pi )={\frac {1}{\ln(\pi )}}

f\left({\frac {9}{4}}\pi \right)=0

It is known the following is true: $1\geq {\frac {\pi }{4}}\geq 0$ . From there, we can directly argue the following:

{\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}

By the intermediate value theorem, if

f(x)

is continuous from

x\in \left[\pi ,{\frac {9}{4}}\pi \right]

, then there exists an

x=\mu

so that

f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )

for

\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]

.

\blacksquare

53. Use the intermediate value theorem to show that there exists a value

x=\Gamma

so that

f(\Gamma )=2

for

f(x)={\sqrt {-2x+4}}+5\ln \left(x^{2}\right)

from

-1\leq x\leq 1

. If you cannot use the intermediate value theorem to show this, explain why.

Notice

f(x)

is not continuous for

x=0

since

\lim _{x\to 0}f(x)

is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.

Notice

f(x)

is not continuous for

x=0

since

\lim _{x\to 0}f(x)

is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.

Solutions

External LinksEdit

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